Question

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Answer 1

Step-by-step explanation:

given equation is

${x}^{2} + 7x + 10 = 0$

it is in the form of

$a {x}^{2} + bx + c = 0$

here , a=1 b=7 c=10

to find nature of roots

∆=b²-4ac

∆=7²-4(1)(10)

∆=49-40

∆=9

as ∆>0

roots are real

now let's find roots...,,,

${x}^{2} + 2x + 5x + 10$

$x(x + 2) + 5(x + 2)$

$(x + 2)(x + 5)$

x+2=0. or x+5=0

x is -2 or -5

Answer 2

Answer:

${x}^{2} + 7x + 10 = 0 \\ = > {x}^{2} + 2x + 5x + 10 = 0 \\ = > x(x + 2) + 5(x + 2) = 0\\ = > (x + 2)(x + 5) = 0 \\ so \: by \: zero \: product \: rule \\ (x + 2) = 0 \: or \: (x + 5) = 0 \\ x + 2 = 0 \\ = > x = - 2 \\ x + 5 = 0 \\ = > x = - 5 \\ so \: the \: roots \: of \: the \: given \: equation \: are \: \\ - 2 \: and \: - 5$

-2 and -5 are the roots of the given equation.

The Explanation is given in the image.

Hope this helps you.